lf the eccentricity of the hyperbola `x^2 - y^2 sec^2 alpha=5` is `sqrt3` times the eccentricity of the ellipse `x^2 (sec^2alpha )+y^2=25,` then a value of `alpha` is : (a) `pi/6` (b) `pi/4` (c) `pi/3` (d) `pi/2`

Options

A. `pi//6`
B. `pi//4`
C. `pi//3`
D. `pi//2`

Explanation

To solve the problem, we need to find the value of \( \alpha \) given the relationship between the eccentricities of a hyperbola and an ellipse. Let's break it down step by step.

### Step 1: Identify the equations of the hyperbola and ellipse.
The hyperbola is given by:
\[ x^2 - y^2 \sec^2 \alpha = 5 \]
The ellipse is given by:
\[ x^2 \sec^2 \alpha + y^2 = 25 \]

### Step 2: Rewrite the ellipse in standard form.
We can rewrite the equation of the ellipse as:
\[ \frac{x^2}{25 \sec^2 \alpha} + \frac{y^2}{25} = 1 \]
This shows that \( a^2 = 25 \sec^2 \alpha \) and \( b^2 = 25 \).

### Step 3: Calculate the eccentricity of the ellipse.
The eccentricity \( e_e \) of the ellipse is given by:
\[ e_e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{25 \sec^2 \alpha}} = \sqrt{1 - \frac{1}{\sec^2 \alpha}} = \sqrt{1 - \cos^2 \alpha} = \sin \alpha \]

### Step 4: Rewrite the hyperbola in standard form.
The hyperbola can be rewritten as:
\[ \frac{x^2}{5} - \frac{y^2}{5 \cos^2 \alpha} = 1 \]
This shows that \( a^2 = 5 \) and \( b^2 = 5 \cos^2 \alpha \).

### Step 5: Calculate the eccentricity of the hyperbola.
The eccentricity \( e_h \) of the hyperbola is given by:
\[ e_h = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5 \cos^2 \alpha}{5}} = \sqrt{1 + \cos^2 \alpha} = \sqrt{2 - \sin^2 \alpha} \]

### Step 6: Set up the relationship between the eccentricities.
According to the problem, the eccentricity of the hyperbola is \( \sqrt{3} \) times the eccentricity of the ellipse:
\[ e_h = \sqrt{3} e_e \]
Substituting the expressions we found:
\[ \sqrt{2 - \sin^2 \alpha} = \sqrt{3} \sin \alpha \]

### Step 7: Square both sides to eliminate the square root.
Squaring both sides gives:
\[ 2 - \sin^2 \alpha = 3 \sin^2 \alpha \]
Rearranging this gives:
\[ 2 = 4 \sin^2 \alpha \]
Thus, we find:
\[ \sin^2 \alpha = \frac{1}{2} \]

### Step 8: Solve for \( \alpha \).
Taking the square root gives:
\[ \sin \alpha = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \alpha = \frac{\pi}{4} \]

### Conclusion:
The value of \( \alpha \) is:
\[ \boxed{\frac{\pi}{4}} \]